| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex numbers 2 |
| Type | Integration using De Moivre identities |
| Difficulty | Challenging +1.2 This is a structured Further Maths question that guides students through standard De Moivre techniques. Parts (a) and (b) are routine algebraic manipulations and theorem applications. Part (c) requires systematic substitution and expansion but follows a well-established method. Part (d) is straightforward integration once part (c) is complete. While it requires multiple steps and is harder than typical A-level pure maths, it's a standard FP2 exercise with clear scaffolding throughout. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.02n Euler's formula: e^(i*theta) = cos(theta) + i*sin(theta)4.02q De Moivre's theorem: multiple angle formulae |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a)(i) \(\left(z + \frac{1}{z}\right)\left(z - \frac{1}{z}\right) = z^2 - \frac{1}{z^2}\) | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \left(z^4 - 2 + \frac{1}{z^4}\right)\left(z^2 + 2 + \frac{1}{z^2}\right)\) | M1A1 | Alternatives for M1A1: \(\left(z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4}\right)\left(z^2 - 2 + \frac{1}{z^2}\right)\) or \(\left(z^3 - \frac{1}{z}\right)^2 - 2\left(z^3 - \frac{1}{z}\right)\left(z - \frac{1}{z}\right) + \left(z - \frac{1}{z}\right)^2\) |
| \(= z^6 + \frac{1}{z^6} + 2\left(z^4 + \frac{1}{z^4}\right) - \left(z^2 + \frac{1}{z^2}\right) - 4\) | A1 | 3 marks; CAO (not necessarily in this form) |
| 8(b)(i) \(z^n + \frac{1}{z^n} = \cos n\theta + i\sin n\theta + \cos(-n\theta) + i\sin(-n\theta) = 2\cos n\theta\) | M1A1 | AG; SC: if solution is incomplete and \((\cos \theta + i\sin \theta)^{-n}\) is written as \(\cos n\theta - i\sin n\theta\), award M1A0A1 |
| 8(b)(ii) \(z^n - z^{-n} = 2i\sin n\theta\) | B1 | 1 mark |
| 8(c) RHS \(= 2\cos 6\theta + 4\cos 4\theta - 2\cos 2\theta - 4\) | M1, A1F |
| Answer | Marks | Guidance |
|---|---|---|
| \(\cos^4 \theta \sin^2 \theta = -\frac{1}{32}\cos 6\theta - \frac{1}{16}\cos 4\theta + \frac{1}{32}\cos 2\theta + \frac{1}{16}\) | M1, A1 | 4 marks; ft incorrect values in (a)(ii) provided they are cosines |
| 8(d) \(\frac{\sin 6\theta}{192} - \frac{\sin 4\theta}{64} + \frac{\sin 2\theta}{64} + \frac{\theta}{16} (+k)\) | M1, A1F | 2 marks; ft incorrect coefficients but not letters A, B, C, D |
**8(a)(i)** $\left(z + \frac{1}{z}\right)\left(z - \frac{1}{z}\right) = z^2 - \frac{1}{z^2}$ | B1 | 1 mark
**8(a)(ii)** $\left(z^2 - \frac{1}{z^2}\right)\left(z + \frac{1}{z}\right)$
$= \left(z^4 - 2 + \frac{1}{z^4}\right)\left(z^2 + 2 + \frac{1}{z^2}\right)$ | M1A1 | Alternatives for M1A1: $\left(z^4 + 4z^2 + 6 + \frac{4}{z^2} + \frac{1}{z^4}\right)\left(z^2 - 2 + \frac{1}{z^2}\right)$ or $\left(z^3 - \frac{1}{z}\right)^2 - 2\left(z^3 - \frac{1}{z}\right)\left(z - \frac{1}{z}\right) + \left(z - \frac{1}{z}\right)^2$
$= z^6 + \frac{1}{z^6} + 2\left(z^4 + \frac{1}{z^4}\right) - \left(z^2 + \frac{1}{z^2}\right) - 4$ | A1 | 3 marks; CAO (not necessarily in this form)
**8(b)(i)** $z^n + \frac{1}{z^n} = \cos n\theta + i\sin n\theta + \cos(-n\theta) + i\sin(-n\theta) = 2\cos n\theta$ | M1A1 | AG; SC: if solution is incomplete and $(\cos \theta + i\sin \theta)^{-n}$ is written as $\cos n\theta - i\sin n\theta$, award M1A0A1
**8(b)(ii)** $z^n - z^{-n} = 2i\sin n\theta$ | B1 | 1 mark
**8(c)** RHS $= 2\cos 6\theta + 4\cos 4\theta - 2\cos 2\theta - 4$ | M1, A1F
LHS $= -64\cos^4 \theta \sin^2 \theta$
$\cos^4 \theta \sin^2 \theta = -\frac{1}{32}\cos 6\theta - \frac{1}{16}\cos 4\theta + \frac{1}{32}\cos 2\theta + \frac{1}{16}$ | M1, A1 | 4 marks; ft incorrect values in (a)(ii) provided they are cosines
**8(d)** $\frac{\sin 6\theta}{192} - \frac{\sin 4\theta}{64} + \frac{\sin 2\theta}{64} + \frac{\theta}{16} (+k)$ | M1, A1F | 2 marks; ft incorrect coefficients but not letters A, B, C, D
**TOTAL MARKS: 75**8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Expand
$$\left( z + \frac { 1 } { z } \right) \left( z - \frac { 1 } { z } \right)$$
\item Hence, or otherwise, expand
$$\left( z + \frac { 1 } { z } \right) ^ { 4 } \left( z - \frac { 1 } { z } \right) ^ { 2 }$$
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use De Moivre's theorem to show that if $z = \cos \theta + \mathrm { i } \sin \theta$ then
$$z ^ { n } + \frac { 1 } { z ^ { n } } = 2 \cos n \theta$$
\item Write down a corresponding result for $z ^ { n } - \frac { 1 } { z ^ { n } }$.
\end{enumerate}\item Hence express $\cos ^ { 4 } \theta \sin ^ { 2 } \theta$ in the form
$$A \cos 6 \theta + B \cos 4 \theta + C \cos 2 \theta + D$$
where $A , B , C$ and $D$ are rational numbers.
\item Find $\int \cos ^ { 4 } \theta \sin ^ { 2 } \theta d \theta$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2008 Q8 [14]}}